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Quizzes > High School Quizzes > Mathematics

Solving Rational Equations Practice Quiz

Boost Confidence with Rational Functions and Equation Solving

Difficulty: Moderate
Grade: Grade 10
Study OutcomesCheat Sheet
Colorful paper art promoting a rational graphs and equations practice quiz for high school students.

What is the domain of the function f(x) = (x + 3)/(x - 2)?
All real numbers except x = 2
All real numbers except x = -3
Only x = 2
All real numbers
The function is undefined when the denominator equals 0. Since x - 2 is 0 when x = 2, that value must be excluded from the domain.
For the rational function f(x) = (2x - 4)/(x + 5), what is the vertical asymptote?
x = 2
x = -5
x = -2
x = 5
The vertical asymptote occurs where the denominator is zero. Setting x + 5 = 0 gives x = -5, which is excluded from the domain.
What does a hole in the graph of a rational function represent?
A removable discontinuity where a common factor cancels
The y-intercept of the function
A vertical asymptote where the function is undefined
A point where the graph changes direction
A hole occurs when a factor cancels from both the numerator and denominator, resulting in a removable discontinuity. It indicates a specific point on the graph where the function is not defined.
Which of the following represents the horizontal asymptote for f(x) = (3x + 1)/(x - 4)?
y = 1
y = 0
y = 3
y = -3
Both the numerator and denominator are linear functions. The horizontal asymptote is determined by dividing the leading coefficients, resulting in y = 3.
What is the sum of the valid solutions of the rational equation (x/2) = (3/(x + 1))?
-1
5
-3
1
Cross-multiplying gives x(x + 1) = 6, which simplifies to a quadratic equation x² + x - 6 = 0. The sum of the roots of this quadratic, by Vieta's formulas, is -1.
For f(x) = (x² - 9)/(x² - x - 12), which statement is true regarding its discontinuities?
There is a hole at x = -3 and a vertical asymptote at x = 4
There is a hole at x = 3 and a vertical asymptote at x = -4
There are vertical asymptotes at x = -3 and x = 4
There is a hole at x = 4 and a vertical asymptote at x = -3
Factoring the numerator gives (x - 3)(x + 3) and the denominator factors to (x - 4)(x + 3). The common factor (x + 3) cancels, producing a hole at x = -3, while the remaining factor (x - 4) in the denominator creates a vertical asymptote at x = 4.
Find the horizontal asymptote of f(x) = (2x² + 3x - 5)/(x² - 4).
No horizontal asymptote
y = 0
y = -2
y = 2
Since the degrees of the numerator and denominator are equal (both quadratic), the horizontal asymptote is the ratio of the leading coefficients, which is 2/1 = 2.
Identify the slant (oblique) asymptote of f(x) = (x² + x + 1)/(x - 1).
y = x + 2
y = x + 1
y = x - 2
y = 2x + 1
Dividing the quadratic numerator by the linear denominator yields a quotient of x + 2, which is the slant asymptote. This occurs because the degree of the numerator is exactly one more than that of the denominator.
Solve the rational equation (2/(x - 3)) + (3/(x + 2)) = (5/(x - 3)).
No solution
x = -2
x = 5
x = 3
After subtracting 2/(x - 3) from both sides, the equation simplifies to 3/(x + 2) = 3/(x - 3). Cross multiplying then leads to a contradiction, meaning there is no valid solution.
For the equation x/(x - 2) = (x + 4)/(x² - 4), which statement about extraneous solutions is true?
x = 2 is an extraneous solution
x = -2 is an extraneous solution
Both x = 2 and x = -2 are extraneous solutions
There are no extraneous solutions for this equation
After cross-multiplying and simplifying, the solutions obtained do not cause any denominator in the original equation to equal zero. Thus, no extraneous solutions are introduced.
Solve for x: (x - 1)/(x + 3) = (2x + 5)/(x² + 2x - 3).
No solution
x = 2 + 2√2
x = 2 + 2√2 and x = 2 - 2√2
x = 1 and x = -3
Factoring the quadratic in the denominator reveals common factors that allow cross-multiplication. Solving the resulting quadratic equation leads to the solutions x = 2 + 2√2 and x = 2 - 2√2.
Find the oblique asymptote of f(x) = (3x² + x - 2)/(x - 1).
y = 4x + 3
y = 3x + 4
y = 3x + 2
y = 2x + 4
Polynomial long division of the numerator by the denominator produces a quotient of 3x + 4. This quotient represents the slant asymptote, as the degree of the numerator is one higher than that of the denominator.
Determine the x-intercepts of f(x) = (x² - 4)/(x² - 9).
x = 2 only
x = 2 and x = -2
x = 3 and x = -3
x = -2 only
The x-intercepts are found by setting the numerator equal to zero, which gives x = 2 or x = -2. Since these values do not make the denominator zero, they are valid intercepts.
For f(x) = (2x)/(x² - 1), what are the vertical asymptotes?
x = 0
x = 1 only
x = 1 and x = -1
x = -1 only
The denominator factors as (x - 1)(x + 1), so the function is undefined at x = 1 and x = -1. These values are the locations of the vertical asymptotes.
Solve the rational equation (x/(x + 4)) = (4/(x - 2)).
x = 8 only
x = 2 and x = -4
x = -2 only
x = 8 and x = -2
Cross multiplying yields x(x - 2) = 4(x + 4), which simplifies to the quadratic equation x² - 6x - 16 = 0. Solving this quadratic gives the solutions x = 8 and x = -2.
For f(x) = (x² - 4)/(x² - 5x + 6), which statement correctly describes its discontinuities?
There are no discontinuities
There is a hole at x = 3 and a vertical asymptote at x = 2
There is a hole at x = 2 and a vertical asymptote at x = 3
Both x = 2 and x = 3 are vertical asymptotes
Factoring the numerator gives (x - 2)(x + 2) and the denominator factors as (x - 2)(x - 3). The common factor (x - 2) cancels, creating a removable discontinuity (hole) at x = 2, while the factor (x - 3) in the denominator establishes a vertical asymptote at x = 3.
Solve the rational equation (2/(x - 1)) + (3/(x + 2)) = (7x + 1)/(x² + x - 2).
x = 1
x = -2
No solution
x = 0
Note that x² + x - 2 factors as (x - 1)(x + 2). Multiplying through by the common denominator and simplifying leads to the equation 5x + 1 = 7x + 1, which yields x = 0. This solution does not violate any restrictions.
Find the y-intercept of f(x) = (3x - 6)/(x² - 4).
y = -3/2
y = 3/2
y = 2
y = 0
To find the y-intercept, substitute x = 0 into the function. This gives f(0) = (-6)/(-4) = 3/2, so the y-intercept is y = 3/2.
Determine the end behavior of f(x) = (5x³ - 2x + 1)/(2x³ + x - 6).
The function approaches y = 0
The function increases without bound
The function oscillates as x approaches ±∞
The function approaches y = 5/2 as x approaches ±∞
Since the degrees of the numerator and the denominator are both 3, the end behavior is determined by the ratio of the leading coefficients. Therefore, as x approaches infinity or negative infinity, f(x) approaches 5/2.
For f(x) = (x² - 9)/(x² - 4x + 3), determine its intercepts, vertical asymptote, and hole.
The function has a hole at x = 1, a vertical asymptote at x = 3, an x-intercept at x = 3, and a y-intercept at (0, -3)
The function has a hole at x = 3, a vertical asymptote at x = -3, an x-intercept at x = 1, and a y-intercept at (0, -3)
The function has no holes, a vertical asymptote at x = 1, an x-intercept at x = -3, and a y-intercept at (0, 3)
The function has a hole at x = 3, a vertical asymptote at x = 1, an x-intercept at x = -3, and a y-intercept at (0, -3)
Factoring shows that the numerator (x² - 9) factors into (x - 3)(x + 3) and the denominator (x² - 4x + 3) factors into (x - 1)(x - 3). The common factor (x - 3) cancels, resulting in a hole at x = 3. The simplified function (x + 3)/(x - 1) reveals a vertical asymptote at x = 1, while the intercepts are determined by setting the numerator to zero and evaluating the function at x = 0.
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Study Outcomes

  1. Analyze rational functions to identify asymptotes, intercepts, and discontinuities.
  2. Graph rational equations accurately by plotting critical points and behavior.
  3. Apply algebraic techniques to solve and simplify rational equations.
  4. Determine the domain and range of rational functions from their equations.
  5. Interpret the effects of transformations on the shapes and positions of graphs.

Rational Functions & Equations Quiz 1 Cheat Sheet

  1. Definition of Rational Functions - Rational functions are quotients of two polynomials, written as f(x) = p(x)/q(x) where q(x) ≠ 0. Recognizing this form is your first step to mastering their quirks and plots. Graphing Rational Functions
  2. Vertical Asymptotes - Find where the denominator equals zero to locate vertical asymptotes - those "invisible walls" where the graph shoots toward ±∞. Understanding these lines helps you sketch the graph's dramatic excursions. Algebra - Rational Functions
  3. Horizontal Asymptotes - Compare the degrees of numerator and denominator to predict end behavior: if they're equal, you get a constant; if one is bigger, the graph tends toward 0 or infinity. This tip turns tail-end mysteries into clear predictions. Algebra - Rational Functions
  4. X‑Intercepts - Set the numerator equal to zero and solve for x to find where the graph crosses the x‑axis. These intercepts are like treasure markers - pin them down to anchor your sketch. Graphing Rational Functions
  5. Y‑Intercepts - Plug in x = 0 and compute f(0) to see where the graph meets the y‑axis. This quick check gives you a solid starting point for your curve. Graphing Rational Functions
  6. Removable Discontinuities (Holes) - When numerator and denominator share a factor you can cancel, you've got a hole - points where the function is undefined but the graph almost has a dot. Spotting holes keeps your plot precise. Study Guide - Rational Functions
  7. Sketching the Graph - Plot intercepts, asymptotes, holes, and extra points to see the full picture. With these landmarks in place, connecting the dots becomes a breeze - and practice makes perfect. Algebra - Practice Problems
  8. Constructing from Conditions - Given specific intercepts or asymptotes, reverse-engineer p(x) and q(x) by building factors that match those features. It's like solving a puzzle where every asymptote and intercept is a clue! Study Guide - Graph Rational Functions
  9. Behavior Near Critical Points - Analyze how the graph approaches asymptotes and crosses intercepts to predict ballooning or gentle curves. Knowing local behavior turns surprises into smooth sailing. Study Guide - Graph Rational Functions
  10. Extra Practice Resources - Reinforce your skills with targeted drills and interactive problems. Regular practice solidifies concepts so you can tackle any rational function with confidence - and a smile! Graphing Rational Functions Practice
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